2. Solving Schrödinger’s Equation

2. Solving Schrödinger’s Equation Superposition • Given a few solutions of Schrödinger’s equation, we can make more of them • Let 1 and 2 be two solutions of Schrödinger’s equation • Let c1 and c2 be any pair of complex numbers • Then the following is also a solution of Schrödinger’s equation • We can generalize this to an arbitrary number of solutions: • We can even generalize to a continuum of solutions:

2A. The Free Schrödinger’s Equation We know a lot of solutions • Consider the free Schrödinger equation: • We know a lot of solutions of this equation: • By combining these, we can make a lot more solutions* • The function c(k) can be (almost) any function of k • This actually includes all possible solutions *The factor of (2)3/2 is arbitrary and inserted here for convenience. In 1D, it would be (2)1/2

We can (in principle) always solve this problem Goal: Given (r,t = 0) = (r), find (r,t) when no potential is present • Set t = 0: • This just says that c(k) is the Fourier transform of (r) • Substitute it in the formula above for (r,t) and we are done • Actually doing the integrals may be difficult

Sample Problem A particle in one dimension with no potential has wave function at t = 0 given by What is the wave function at arbitrary time? • Need 1D versions of these formulas: From Appendix: Find c(k):

Sample Problem (2) A particle in one dimension with no potential has wave function at t = 0 given by What is the wave function at arbitrary time? • Now find (x,t):

2B. The Time Independent Schrödinger Eqn. Separation of Variables • Suppose that the potential is independent of time: • Conjecture solutions of the form: • Substitute it in • Divide by (r)(t) • Left side is independent of r • Right side is independent of t • Both sides are independent of both! Must be a constant. Call it E

Solving the time equation • The first equation is easy to solve • Integrate both sides • By comparison with e-it, we see that E = is the energy • Substitute it back in:

The Time Independent Schrödinger Equation • Multiply the other equation by (r) again: The Strategy for solving • Given a potential V(r) independent of time, what is most general solution of Schrödinger’s time-dependent equation? • First, solve Schrödinger’s time-independent equation • You should find many solutions n(r) with different energies En • Now just multiply by the phase factor • Then take linear combinations • Later we’ll learn how to find cn

Why is time-independent better? • Time-independent is one less variable – significantly easier • It is a real equation (in this case), which is less hassle to solve • If in one dimension, it reduces to an ordinary differential equation • These are much easier to solve, especially numerically

2C. Probability current Probability Conservation • Recall the probability density is: • This can change as the wave function changes • Where does the probability “go” as it changes? • Does it “flow” just like electric charge does? • Want to show that the probability moves around • Ideally, show that it “flows” from place to place • A formula from E and M – can we make it like this? • To make things easier, let’s make all functions of space and time implicit, not write them

The derivation (1) • Start with Schrödinger’s equation • Multiply on the left by *: • Take complex conjugate of this equation: • Subtract: • Rewrite first term as a total derivative • Cancel a factor of i • Left side is probability density

The derivation (2) • Consider the following expression: • Use product rule on the divergence • Substitute this in above • Define the probability currentj: • Then we have:

Why is it called probability current? • Integrate it over any volume V with surface S • Left side is P(r V) • Use Gauss’s law on right side • Change in probability is due to current flowing out V j • If the wave function falls off at infinity (as it must) and the volume V becomes all of space, we have

Calculating probability current • This expression is best when doing proofs • Note that you have a real number minus its complex conjugate • A quicker formula for calculation is: • Let’s find  and j for a plane wave:

Sample Problem A particle in the 1D infinite square well has wave function For the region 0 < x < a. Find  and j.

Sample Problem (2) A particle in the 1D infinite square well has wave function For the region 0 < x < a. Find  and j. • In 1D:

Sample Problem (3) A particle in the 1D infinite square well has wave function For the region 0 < x < a. Find  and j. • After some work …

2D. Reflection from a Step Boundary The Case E > V0: Solutions in Each Region I • A particle with energy E impacts a step-function barrier from the left: incident transmitted reflected II Solve the equation in each of the regions • Assume E > V0 • Region I • Region II • Most general solution: • A is incident wave • B is reflected wave • C is transmitted wave • D is incoming wave from the right: set D = 0

Step with E > V0: The solution I incident transmitted reflected • Schrödinger’s equation: second derivative finite • (x) and ’(x) must be continuous at x = 0 II • We can’t normalize wave functions • Use probability currents!

Summary: Step with E > V0 I incident transmitted reflected II

Step with E < V0 • What if V0 > E? • Region I same as before • Region II: we have I II incident reflected evanescent • Most general solution: • A is incident wave • B is reflected wave • C is damped “evanescent” wave • D is growing wave, can’t be normalized • (x) and ’(x) must be continuous at x = 0: • No transmission since evanescent wave is damped

Step Potential: All cases summarized T R • For V0 > E, all is reflected • Note that it penetrates, a little bit into the classically forbidden region, x > 0 • This suggests if barrier had finite thickness, some of it would bet through • Reflection probability:

2E. Quantum Tunneling Setting Up the Problem V(x) • Barrier of finite height and width: V0 • Solve the equation in each of the regions • Particle impacts from left with E < V0 • General solution in all three regions: III II I - d/2 + d/2 x • Match  and ’ at x = -d/2 and x = d/2 Why didn’t I include e-ikx in III? Why did I skip letter E? • Solve for F in terms of A

Skip this Slide – Solving for F in terms of A • Multiply 1 by ik and add to 2 • Multiply 3 by  and add to 4 • Multiply 3 by  and subtract 4 • Multiply 5 by 2 and substitute from 6 and 7

Barrier Penetration Results • We want to know transmission probability • For thick barriers, • Exponential suppression of barrier penetration

Unbound and Bound State • For each of the following, we found solutions for any E • No potential • Step potential • Barrier • This is because we are dealing with unbound states, E > V() • Our wave functions were, in each case, not normalizable • Fixable by making superpositions: • We will now consider bounds states • These are when E < V() • There will always only be discrete energy values • And they can be normalized • Usually easier to deal with real wave functions

2F. The Infinite Square Well Finding the Modes V(x) • Infinite potential implies wave function must vanish there • In the allowed region, Schrödinger’s equation is just x 0 a • The solution to this is simple: • Because potential is infinite, the derivative is not necessarily continuous • But wave functions must still be continuous:

Normalizing Modes and Quantized Energies • We can normalize this wave function: • Note that we only get discrete energies in this case • Note that we can normalize these • Most general solution is then

The 3D Infinite Square Well • In allowed region: • Guess solution: • Normalize it: • This is product of 1D functions • Energy is • This is sum of 1D energies c b a

2G. The Double Delta-Function Potential Finding Bound States V(x) -a/2 a/2 I III II • First, write out Schrödinger’s Equation: x • Bound states have E < V() = 0 • Within each region we have: • General solution (deleting the parts that blow up at infinity):

Dealing with Delta Functions V(x) -a/2 a/2 I III II • To deal with the delta functions, integrate Schrödinger’s equation over a small region near the delta function: • For example, near x = +a/2 • Do first term on right by fundamental theorem of calculus • Do second term on right by using the delta functions • Take the limit   0 • Left side small in this limit

Simplifying at x = ½a V(x) -a/2 a/2 I III II • Since there is a finite discontinuity in ’,  must be continuous at this boundary On the right side of the equation above, is that I, II, or III? • Write these equation out explicitly: • Substitute first into second:

Repeating at x = – ½a • Repeat the steps we did, this time at x = –½a • Note these equations are nearly identical: • The only numbers equal to their reciprocal are 1

Graphical Solution Right side, plus Right side, minus Left side • Right side is two curves, left side is a straight line • Black line always crosses red curve, sometimes crosses green curve, depending on parameters • Sometimes two solutions, sometimes one • Normalize to finish the problem • Note one solution symmetric, one anti-symmetric

2. Solving Schrödinger’s Equation